478 Higher Engineering Mathematics
50.2 Procedure to solve differential
equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
(a) Rewrite the differential equation
a
d 2 y
dx 2 + b
d y
dx
+ cy = 0
as (aD 2 + bD + c)y = 0
(b) Substitute m for D and solve the auxiliary equation
am 2 + bm + c = 0 for m.
(c) If the roots of the auxiliary equation are:
(i) real and different, say m = α and m = β,
then the general solution is
y = Ae
αx
+ Be
βx
(ii) real and equal, say m = α twice, then the
general solution is
y = (Ax + B)e
αx
(iii) complex, say m = α ± jβ, then the general
solution is
y = e
αx {A cosβx + B sinβx}
(d) Given boundary conditions, constants A and B,
may be determined and the particular solution
of the differential equation obtained.
The particular solutions obtained in the worked problems of Section 50.3 may each be verified by substituting expressions for y,
d y
dx
and
d 2 y
dx 2 into the original
equation.
50.3 Worked problems on
differential equations of
the form a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
Problem 1. Determine the general solution of
2
d 2 y
dx 2 + 5
d y
dx
− 3y = 0. Find also the particular
solution given that when x = 0, y = 4 and
d y
dx
= 9.
Using the above procedure:
(a) 2
d 2 y
dx 2 + 5
d y
dx
− 3y = 0 in D-operator form is
(2D 2 + 5D − 3)y = 0, where D ≡
d
dx
(b) Substituting m for D gives the auxiliary equation
2m
2
+ 5m − 3 = 0.
Factorising gives: (2m − 1)(m + 3) = 0, from
which, m =
1
2 or m =−3.
(c) Since the roots are real and different the general
solution is y = Ae
1
2 x
+ Be −3x .
(d) When x = 0, y = 4,
hence
4 = A + B
(1)
Since
y = Ae
1
2 x
+ Be −3x
then
d y
dx
=
1
2
Ae
1
2 x
− 3Be −3x
When
x = 0,
d y
dx
= 9
thus
9 =
1
2
A − 3B
(2)
Solving the simultaneous equations (1) and (2)
gives A = 6 and B =−2.
Hence the particular solution is
y = 6e
1
2 x
− 2e
−3x
Problem 2. Find the general solution of
9
d 2 y
dt 2 − 24
d y
dt
+ 16y = 0 and also the particular
solution given the boundary conditions that when
t = 0, y =
d y
dt
= 3.
Using the procedure of Section 50.2:
(a) 9
d 2 y
dt 2 − 24
d y
dt
+ 16y = 0 in D-operator form is
(9D 2 − 24D +16)y = 0 where D ≡
d
dt
(b) Substituting m for D gives the auxiliary equation
9m 2 − 24m + 16 =0.
Factorizing gives: (3m − 4)(3m − 4) = 0, i.e.
m =
4
3 twice.
(c) Since the roots are real and equal, the general
solution is y = (At +B)e
4
3 t .
50.2 Procedure to solve differential
equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
(a) Rewrite the differential equation
a
d 2 y
dx 2 + b
d y
dx
+ cy = 0
as (aD 2 + bD + c)y = 0
(b) Substitute m for D and solve the auxiliary equation
am 2 + bm + c = 0 for m.
(c) If the roots of the auxiliary equation are:
(i) real and different, say m = α and m = β,
then the general solution is
y = Ae
αx
+ Be
βx
(ii) real and equal, say m = α twice, then the
general solution is
y = (Ax + B)e
αx
(iii) complex, say m = α ± jβ, then the general
solution is
y = e
αx {A cosβx + B sinβx}
(d) Given boundary conditions, constants A and B,
may be determined and the particular solution
of the differential equation obtained.
The particular solutions obtained in the worked problems of Section 50.3 may each be verified by substituting expressions for y,
d y
dx
and
d 2 y
dx 2 into the original
equation.
50.3 Worked problems on
differential equations of
the form a
d 2 y
dx 2 + b
dy
dx
+ cy = 0
Problem 1. Determine the general solution of
2
d 2 y
dx 2 + 5
d y
dx
− 3y = 0. Find also the particular
solution given that when x = 0, y = 4 and
d y
dx
= 9.
Using the above procedure:
(a) 2
d 2 y
dx 2 + 5
d y
dx
− 3y = 0 in D-operator form is
(2D 2 + 5D − 3)y = 0, where D ≡
d
dx
(b) Substituting m for D gives the auxiliary equation
2m
2
+ 5m − 3 = 0.
Factorising gives: (2m − 1)(m + 3) = 0, from
which, m =
1
2 or m =−3.
(c) Since the roots are real and different the general
solution is y = Ae
1
2 x
+ Be −3x .
(d) When x = 0, y = 4,
hence
4 = A + B
(1)
Since
y = Ae
1
2 x
+ Be −3x
then
d y
dx
=
1
2
Ae
1
2 x
− 3Be −3x
When
x = 0,
d y
dx
= 9
thus
9 =
1
2
A − 3B
(2)
Solving the simultaneous equations (1) and (2)
gives A = 6 and B =−2.
Hence the particular solution is
y = 6e
1
2 x
− 2e
−3x
Problem 2. Find the general solution of
9
d 2 y
dt 2 − 24
d y
dt
+ 16y = 0 and also the particular
solution given the boundary conditions that when
t = 0, y =
d y
dt
= 3.
Using the procedure of Section 50.2:
(a) 9
d 2 y
dt 2 − 24
d y
dt
+ 16y = 0 in D-operator form is
(9D 2 − 24D +16)y = 0 where D ≡
d
dt
(b) Substituting m for D gives the auxiliary equation
9m 2 − 24m + 16 =0.
Factorizing gives: (3m − 4)(3m − 4) = 0, i.e.
m =
4
3 twice.
(c) Since the roots are real and equal, the general
solution is y = (At +B)e
4
3 t .